TG EAPCET 2022 20th July Morning Shift Question Paper with Solutions

$\begin{array}{cc}& \text{ Given, }𝑓\left(𝑥\right)=\sqrt{\surd \frac{2{𝑥}^2-7𝑥+5}{3{𝑥}^2-5𝑥-2}}\\ & \frac{2{𝑥}^2-7𝑥+5}{3{𝑥}^2-5𝑥-2}\ge 0\\ & \Rightarrow \frac{2{𝑥}^2-2𝑥-5𝑥+5}{3{𝑥}^2-6𝑥+𝑥-2}\ge 0\\ & \Rightarrow \frac{2𝑥(𝑥-1)-5(𝑥-1)}{3𝑥(𝑥-2)+1(𝑥-2)}\ge 0\\ & \Rightarrow \frac{(2𝑥-5)(𝑥-1)}{(3𝑥+1)(𝑥-2)}\ge 0\\ & \Rightarrow 𝑥\ne -\frac{1}{3},𝑥\ne 2\end{array}$

$\begin{array}{cc}& \text{ Domain : }\\ & 𝑥\in (-\infty ,-\frac{1}{3})\cup [1,2)\cup [\frac{5}{2},\infty )\end{array}$

Given, $𝑓\left(𝑥\right)=|𝑥-2|+|𝑥-3|$

We know that for $𝑓\left(𝑥\right)=|𝑥-𝑎|+|𝑥-𝑏|$, the minimum value of $𝑓\left(𝑥\right)$,

we get either at $𝑥=𝑎$

or at $𝑥=𝑏$

$\begin{array}{cc}𝑆𝑜𝑓𝑜𝑟& 𝑓\left(𝑥\right)=|𝑥-2|+|𝑥-3|\\ & 𝑓\left(2\right)=0+1=1\\ & 𝑓\left(3\right)=1+0=1\end{array}$

So, $𝑓(𝑥{)}_{\text{min }}=1$ and maximum value will be $\infty$ at $𝑥=\infty$.

$\because {𝑓}_{max}=\infty +\infty =\infty$

So, range $=[1,\infty )$

             $\begin{array}{cc}& \text{ Given, }𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 3\\ 4& 3& 2\\ 3& 4& 5\end{array}⎤⎦⎥\text{ then, }\\ & {𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}1& 4& 3\\ 2& 3& 4\\ 3& 2& 5\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}(𝐴+{𝐴}^{𝑇})& (𝐴-{𝐴}^{𝑇})\\ & =⎡⎣⎢\begin{array}{ccc}2& 6& 6\\ 6& 6& 6\\ 6& 6& 10\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}0& -2& 0\\ 2& 0& -2\\ 0& 2& 0\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}12& 8& -12\\ 12& 0& -12\\ 12& 8& -12\end{array}⎤⎦⎥\\ & =4⎡⎣⎢\begin{array}{ccc}3& 2& -3\\ 3& 0& -3\\ 3& 2& -3\end{array}⎤⎦⎥\end{array}$

Given,

$𝑓\left(𝑥\right)=\mid \begin{array}{ccc}𝑥& 𝑥+1& 𝑥+3\\ 𝑥+2& 𝑥+4& 𝑥+7\\ 𝑥+6& 𝑥+9& 𝑥+13\end{array}\mid$

Applying ${𝑅}_3\to {𝑅}_3-{𝑅}_1,{𝑅}_2\to {𝑅}_2-{𝑅}_1$,

$𝑓\left(𝑥\right)=\mid \begin{array}{ccc}𝑥& 𝑥+1& 𝑥+3\\ 2& 3& 4\\ 6& 8& 10\end{array}\mid$

$\begin{array}{cc}& \mathrm{At}𝑥=5,𝑓\left(5\right)=\mid \begin{array}{ccc}5& 6& 8\\ 2& 3& 4\\ 6& 8& 10\end{array}\mid \\ & =5(30-32)-6(20-24)+8(16-18)\\ & =-10+24-16=-2\end{array}$

Given, $𝐴=⎡⎣⎢\begin{array}{ccc}2& 1& 1\\ 0& 1& 0\\ 1& 1& 2\end{array}⎤⎦⎥$

$\because {𝐴}^{-1}=𝛼{𝐴}^2+𝛽𝐴+𝛾𝐼$

Multiply by $𝐴$ both sides,

$\begin{array}{cc}{𝐴}^{-1}\cdot 𝐴& =𝛼{𝐴}^3+𝛽{𝐴}^2+𝛾𝐼\cdot 𝐴\\ 𝐼& =𝛼{𝐴}^3+𝛽{𝐴}^2+𝛾𝐴\end{array}$

For matrix $𝐴$

$\det .(𝐴-𝜆𝐼)=0$

$\mid \begin{array}{ccc}2-𝜆& 1& 1\\ 0& 1-𝜆& 0\\ 1& 1& 2-𝜆\end{array}\mid =0$

$\text{ Expanding along }{𝑅}_1$

$\begin{array}{cc}& (2-𝜆)(1-𝜆)(2-𝜆)-1\left(0\right)+1(𝜆-1)=0\\ & \Rightarrow (1-𝜆)({𝜆}^2-4𝜆+4)+𝜆-1=0\\ & \Rightarrow {𝜆}^2-4𝜆+4-{𝜆}^3+4{𝜆}^2-4𝜆+𝜆-1=0\end{array}$

$\begin{array}{cc}\Rightarrow & -{𝜆}^3+5{𝜆}^2-7𝜆+3=0\\ \Rightarrow & \frac{1}{3}{𝜆}^3-\frac{5}{3}{𝜆}^2+\frac{7}{3}𝜆=1\end{array}$

$\begin{array}{cc}& \text{ Compare with }𝛼{𝐴}^3+𝛽{𝐴}^2+𝛾𝐴=𝐼\text{, }\\ & \begin{array}{cc}& \text{ So, }𝛼=\frac{1}{3},𝛽=\frac{-5}{3},𝛾=\frac{7}{3}\\ & \therefore 17𝛼+5𝛽+𝛾=\frac{17}{3}-\frac{25}{3}+\frac{7}{3}=-\frac{1}{3}\end{array}\end{array}$

Here,

$𝐴𝑋=⎡⎣⎢\begin{array}{c}1\\ 1\\ 2\end{array}⎤⎦⎥,\mathrm{Adj}𝐴=⎡⎣⎢\begin{array}{ccc}1& -1& -1\\ 1& 1& -1\\ 1& 1& 1\end{array}⎤⎦⎥$ and also $\left|𝐴\right|>0$.

Now, using $|\mathrm{adj}𝐴|=|𝐴{|}^{𝑛-1};𝑛$ is order of square matrix $𝐴$.

$\begin{array}{cc}& \mid \begin{array}{ccc}1& -1& -1\\ 1& 1& -1\\ 1& 1& 1\end{array}\mid =|𝐴{|}^{3-1}\\ & \Rightarrow |𝐴{|}^2=1(2)+1(2)-1(0)=4\\ & \Rightarrow \left|𝐴\right|=2[\because |𝐴|>0]\end{array}$

$\begin{array}{cc}𝑁𝑜𝑤,𝑋& ={𝐴}^{-1}⎡⎣⎢\begin{array}{c}1\\ 1\\ 2\end{array}⎤⎦⎥=\frac{\mathrm{Adj}\left(𝐴\right)}{\left|𝐴\right|}⎡⎣⎢\begin{array}{c}1\\ 1\\ 2\end{array}⎤⎦⎥\\ & =\frac{1}{2}⎡⎣⎢\begin{array}{ccc}1& -1& -1\\ 1& 1& -1\\ 1& 1& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}1\\ 1\\ 2\end{array}⎤⎦⎥\\ \Rightarrow 𝑋& =\frac{1}{2}⎡⎣⎢\begin{array}{c}-2\\ 0\\ 4\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}-1\\ 0\\ 2\end{array}⎤⎦⎥\end{array}$

For $𝑥\in [0,2𝜋]$

$\sin 𝑥+𝑖\cos 2𝑥$ and $\cos 𝑥-𝑖\sin 2𝑥$ are conjugate to each other.

$\begin{array}{cc}\Rightarrow \sin 𝑥+𝑖\cos 2𝑥& =\overline{\cos 𝑥-𝑖\sin 2𝑥}\\ \sin 𝑥+𝑖\cos 2𝑥& =\cos 𝑥+𝑖\sin 2𝑥...\left(𝑖\right)\end{array}$

On comparing both side of Eq. (i), we get

$\begin{array}{ccccc}& & \sin 𝑥& =\cos 𝑥;& \sin 2𝑥=\cos 2𝑥\\ \Rightarrow & & 𝑥& =\frac{𝜋}{4},\frac{5𝜋}{4};& \tan 2𝑥=1\\ \Rightarrow & & 2𝑥& =\frac{𝜋}{4},\frac{5𝜋}{4},\frac{9𝜋}{4},\frac{13𝜋}{4}\\ \Rightarrow & & 𝑥& =\frac{𝜋}{8},\frac{5𝜋}{8},\frac{9𝜋}{8},\frac{13𝜋}{8}\end{array}$

But there is no common value of $𝑥$ which satisfy both the condition. So, $𝑥=𝜙$

Given, $|𝑥+𝑖𝑦|=\sqrt{\surd {𝑥}^2+{𝑦}^2}$

$\begin{array}{cc}𝑆𝑖𝑛𝑐𝑒,& (1-\sqrt{\surd 3}𝑖{)}^9+(\sqrt{\surd 3}+𝑖{)}^9\\ & ={𝑖}^{-9}{(\frac{1}{𝑖}-\sqrt{\surd 3})}^9+(\sqrt{\surd 3}+𝑖{)}^9\\ & [\because {𝑖}^{-9}=\frac{1}{{𝑖}^9}=\frac{1}{{𝑖}^8\cdot 𝑖}=\frac{1}{{\left({𝑖}^2\right)}^4𝑖}=\frac{1}{(-1{)}^4𝑖}=\frac{1}{𝑖}=-𝑖]\end{array}$

$\begin{array}{cc}𝑁𝑜𝑤,& \mid (1-\sqrt{\surd 3}𝑖{)}^9+(\sqrt{\surd 3}+𝑖{)}^9\mid \\ & =\mid (𝑖+\sqrt{\surd 3}{)}^9(𝑖+1)\mid \\ & =|𝑖+1|\mid (𝑖+\sqrt{\surd 3}{)}^9\mid \\ & =|𝑖+1||𝑖+\sqrt{\surd 3}{|}^9\\ & ={\left(\sqrt{\surd 1^2+1^2}\right)}^{{\left(\sqrt{\surd 1^2+(\sqrt{\surd 3}{)}^2}\right)}^9}\\ & =\sqrt{\surd 2}(\sqrt{\surd 4}{)}^9\\ & =2^{\frac{1}{2}}\cdot 2^9=2^{\frac{19}{2}}\end{array}$

If ${𝑥}^3=1$

Then, $𝑥=1,𝜔,{𝜔}^2$ and ${𝑥}^4=1$

$\begin{array}{cc}\Rightarrow & {𝑥}^4-1=0\\ \Rightarrow & 𝑥\pm 1,\pm 𝑖\end{array}$

So, roots are $1,𝑖,-1,-𝑖$

i.e. $𝛼=𝑖$

$[\because 1,𝛼,{𝛼}^2,{𝛼}^3$ are fourth roots of unity]

Now, $𝛼+𝛼𝜔-{𝛼}^3{𝜔}^2=𝛼+𝛼𝜔-\frac{{𝜔}^2}{𝛼}$

$\begin{array}{cc}[\because 𝛼=𝑖\Rightarrow {𝛼}^3={𝑖}^3& \Rightarrow {𝛼}^3={𝑖}^2\cdot 𝑖\\ & \Rightarrow {𝛼}^3=-𝑖\Rightarrow {𝛼}^3=-𝛼]\\ =& 𝛼(1+𝜔)-\frac{{𝜔}^2}{𝛼}=-𝛼{𝜔}^2-\frac{{𝜔}^2}{𝛼}\\ =& -{𝜔}^2(𝛼+\frac{1}{𝛼})=-{𝜔}^2\left(\frac{{𝛼}^2+1}{𝛼}\right)\\ =& -{𝜔}^2\frac{(-1+1)}{𝑖}=0\end{array}$

Given, equation ${𝑥}^2+𝑏𝑥+𝑐=0$ has roots $𝛼$ and $𝛽$, then

$𝛼+𝛽=-𝑏,𝛼𝛽=𝑐$

Given, ${𝛼}^2+{𝛽}^2=5\Rightarrow (𝛼+𝛽{)}^2-2𝛼𝛽=5$

$\Rightarrow {𝑏}^2-2𝑐=5...\left(𝑖\right)$

Also, ${𝛼}^3+{𝛽}^3=9$

$\begin{array}{c}\Rightarrow (𝛼+𝛽)({𝛼}^2+{𝛽}^2-𝛼𝛽)=9\\ \Rightarrow -𝑏(5-𝑐)=9...\left(𝑖𝑖\right)\end{array}$

From Eq. (i), $𝑐=\frac{{𝑏}^2-5}{2}$ on putting in Eq. (ii),

$\begin{array}{cc}& -𝑏(5-\frac{{𝑏}^2-5}{2})=9\\ \Rightarrow & -𝑏\left(\frac{15-{𝑏}^2}{2}\right)=9\\ & \Rightarrow {𝑏}^3-15𝑏-18& =0\end{array}$

Since, $𝑏=-3$ is root of the above equation.

$\begin{array}{cc}& \therefore (𝑏+3)({𝑏}^2-3𝑏-6)=0\\ & 𝑏=-3,𝑐=\frac{{𝑏}^2-5}{2}=\frac{9-5}{2}=2\\ & \therefore 𝑏+𝑐=-1\end{array}$

Let

$\begin{array}{cc}& 𝑦=\frac{{𝑥}^2-𝑥+2}{{𝑥}^2+𝑥-2};\forall 𝑥\in 𝑅-\{-2,1\}\\ \Rightarrow & 𝑦{𝑥}^2+𝑦𝑥-2𝑦={𝑥}^2-𝑥+2\\ \Rightarrow & (𝑦-1){𝑥}^2+(𝑦+1)𝑥-2(𝑦+1)=0\end{array}$

For $𝑥\in 𝑅$,

$\begin{array}{cc}& 𝐷\ge 0\\ & \Rightarrow (𝑦+1{)}^2-4(𝑦-1\left)\right[-2(𝑦+1)]\ge 0\\ & \Rightarrow (𝑦+1)\left[\right(𝑦+1)+8(𝑦-1\left)\right]\ge 0\\ & \Rightarrow (𝑦+1)(9𝑦-7)\ge 0\end{array}$

$\therefore 𝑦\in (-\infty ,-1]\cup [\frac{7}{9},\infty )$

Given, $𝛼,𝛽$ and $𝛾$ are the roots of the equation,

${𝑥}^3-9{𝑥}^2+23𝑥-15=0$

Then, $𝛼+𝛽+𝛾=9$

$\begin{array}{cc}\Rightarrow 𝛼𝛽+𝛽𝛾+𝛾𝛼& =23\\ 𝛼𝛽𝛾& =15\end{array}$

Since,

$\begin{array}{c}{𝛼}^3+{𝛽}^3+{𝛾}^3-3𝛼𝛽𝛾=(𝛼+𝛽+𝛾)\\ \{{𝛼}^2+{𝛽}^2+{𝛾}^2-(𝛼𝛽+𝛽𝛾+𝛾𝛼\left)\right\}\\ \Rightarrow {𝛼}^3+{𝛽}^3+{𝛾}^3=3𝛼𝛽𝛾\\ +(𝛼+𝛽+𝛾)\left[\right(𝛼+𝛽+𝛾{)}^2\\ -3(𝛼𝛽+𝛽𝛾+𝛾𝛼)]\\ =(3\times 15)+9\left\{\right(9{)}^2-3\times 23\}\\ =45+9\times (81-69)=153\end{array}$

Given, $𝛼,𝛽$ and $2𝛽$ are the real roots of the equation ${𝑥}^3-9{𝑥}^2+𝑘=0$ where $𝑘\in 𝑅-\left\{0\right\}$.

Then, $𝛼+𝛽+2𝛽=9$

$\begin{array}{cc}\Rightarrow & 𝛼+3𝛽=9...\left(𝑖\right)\\ \text{ and }& 𝛼\cdot 𝛽\cdot 2𝛽=-𝑘\\ \Rightarrow & 2𝛼{𝛽}^2=-𝑘...\left(𝑖𝑖\right)\end{array}$

Product of two roots

$\begin{array}{c}𝛼𝛽+𝛽\cdot 2𝛽+2𝛽\cdot 𝛼=0\\ \Rightarrow 3𝛼𝛽+2{𝛽}^2=0\\ \Rightarrow 𝛽(3𝛼+2𝛽)=0\\ 𝛽\ne 0\text{ because }𝑘\ne 0\text{ so, }\\ 3𝛼+2𝛽=0& ...\left(𝑖𝑖𝑖\right)\end{array}$

From Eqs. (i) and (iii),

$7𝛽=27\Rightarrow 14𝛽=54$

Given equation,

$\begin{array}{c}{𝑥}^5-3{𝑥}^4+5{𝑥}^3-5{𝑥}^2+3𝑥-1=0\\ \Rightarrow (𝑥-1)({𝑥}^4-2{𝑥}^3+3{𝑥}^2-2𝑥+1)=0\end{array}$

[ $\because$ ] is a root of given

$\begin{array}{c}\Rightarrow (𝑥-1){𝑥}^2({𝑥}^2-2𝑥+3-\frac{2}{𝑥}+\frac{1}{{𝑥}^2})=0\\ \Rightarrow (𝑥-1){𝑥}^2({𝑥}^2+\frac{1}{{𝑥}^2}-2(𝑥+\frac{1}{𝑥})+3)=0\\ (𝑥-1){𝑥}^2\{{(𝑥+\frac{1}{𝑥})}^2-2-2(𝑥+\frac{1}{𝑥})+3\}=0\\ =(𝑥-1){𝑥}^2\{{(𝑥+\frac{1}{𝑥})}^2-2(𝑥+\frac{1}{𝑥})+1\}=0\end{array}$

$\begin{array}{cc}& \Rightarrow (𝑥-1){𝑥}^2{(𝑥+\frac{1}{𝑥}-1)}^2=0\\ & \Rightarrow (𝑥-1){({𝑥}^2-𝑥+1)}^2=0\end{array}$

Now, $𝑥-1=0\Rightarrow 𝑥=1$

$\begin{array}{cc}𝑎𝑛𝑑{𝑥}^2-𝑥+1& =0\\ \Rightarrow 𝑥& =\frac{1\pm 2}{2}=\frac{3}{2},-\frac{1}{2}\end{array}$

So, sum of distinct roots

$=1+\frac{3}{2}-\frac{1}{2}=1+1=2$

Here, $({𝑥}^4+1)=\frac{1}{𝑎}(𝑥+1{)}^4...(𝑖)$

Given equation is a reciprocal equation. Then, replace $𝑥$ by $1/𝑥;𝑥\ne 0$

$\begin{array}{cc}& (\frac{1}{{𝑥}^4}+1)=\frac{1}{𝑎}{(\frac{1}{𝑥}+1)}^4\\ \Rightarrow & \left(\frac{1+{𝑥}^4}{{𝑥}^4}\right)=\frac{1}{𝑎}\frac{(1+𝑥{)}^4}{{𝑥}^4}\\ \Rightarrow & (1+{𝑥}^4)=\frac{1}{𝑎}(1+𝑥{)}^4...(𝑖𝑖)\end{array}$

At $𝑥=0,𝑎=1$

As $𝑥\ne 0,𝑎\in 𝑅-\left\{1\right\}$.

Given, $𝑎+𝑏+𝑐=5$ and $𝑎,𝑏,𝑐\in 𝑁𝐿$ is least value of $2^{𝑎}{3}^{𝑏}{5}^{𝑐}$ and $𝑀$ is the greatest value of $2^{𝑎}{3}^{𝑏}{5}^5$.

Possible pairs of $(𝑎,𝑏,𝑐)$

$\begin{array}{c}=(1,1,3),(1,3,1),(3,1,1),(1,2,2)\\ (2,1,2),(2,2,1)\end{array}$

Value for $(1,1,3)=(2\cdot 3\cdot 5^3)=750$

Value for $(1,3,1)=(2\cdot 3^3\cdot 5)=270$

Value for $(3,1,1)=(2^3\cdot 3\cdot 5)=120$

Value for $(1,2,2)=(2^1\cdot 3^2\cdot 5^2)=450$

Value for $(2,1,2)=(2^2\cdot 3^1\cdot 5^2)=300$

Value for $(2,2,1)=(2^2\cdot 3^2\cdot 5^1)=180$

$\begin{array}{cc}𝐿=2^3\cdot 3\cdot 5& =120,𝑀=2\cdot 3\cdot 5^3=750\\ \therefore 𝑀-𝐿& =2\cdot 3\cdot 5^3-2^3\cdot 3\cdot 5\\ & =2\cdot 3\cdot 5(25-4)\\ & =2\cdot 3\cdot 5\cdot 21=2\cdot 3^2\cdot 5\cdot 7\end{array}$

Since, $360=2^3\cdot 3^2\cdot 5^1$

The number of $(+\mathrm{ve})$ divisors which are multiple of 3

$=(3+1)(2+1-1)(1+1)=4\times 2\times 2=16$

Given,

$\begin{array}{cc}& (\frac{1}{8}-\frac{7}{8\times 12}+\frac{7\times 10}{8\times 12\times 16}-\dots )\\ & =(\frac{3}{4}+\frac{1}{8}-\frac{7}{8\times 12}+\frac{7\times 10}{8\times 12\times 16}-\dots )-\frac{3}{4}\\ & =(1-\frac{1}{4}+\frac{1}{8}-\frac{7}{8\times 12}+\frac{7\times 10}{8\times 12\times 16}-\dots )-\frac{3}{4}\end{array}$

$=⎛⎝⎜\begin{array}{c}1-\frac{1}{3}\times \frac{3}{4}+\frac{\frac{1}{3}\times \frac{4}{3}\times {\left(\frac{3}{4}\right)}^2}{2!}\\ -\frac{\frac{1}{3}\times \frac{4}{3}\times \frac{7}{3}\times {\left(\frac{3}{4}\right)}^3}{3!}\end{array}$

$+\frac{\frac{1}{3}\times \frac{4}{3}\times \frac{7}{3}\times \frac{10}{3}\times {\left(\frac{3}{4}\right)}^4}{4!}-\dots )-\frac{3}{4}$

$=⎛⎝⎜\begin{array}{c}1-\frac{1}{3}\times \frac{3}{4}+\frac{\frac{1}{3}\times (\frac{1}{3}+1)\times {\left(\frac{3}{4}\right)}^2}{2!}\\ -\frac{\frac{1}{3}\times (\frac{1}{3}+1)(\frac{1}{3}+2)\times {\left(\frac{3}{4}\right)}^3}{3!}\end{array}$

$+\frac{\frac{1}{3}(\frac{1}{3}+1)(\frac{1}{3}+2)(\frac{1}{3}+3)\times {\left(\frac{3}{4}\right)}^4}{4!}-\dots )-\frac{3}{4}$

$\begin{array}{cc}& ={(1+\frac{3}{4})}^{-1/3}-\frac{3}{4}\\ & ={\left(\frac{7}{4}\right)}^{-1/3}-\frac{3}{4}=\text{exception 25:}-\frac{3}{4}\end{array}$

Given,

$\begin{array}{cc}& \frac{2{𝑥}^2-3𝑥+5}{(𝑥-7{)}^3}=\frac{𝐴}{𝑥-7}+\frac{𝐵}{(𝑥-7{)}^2}+\frac{𝐶}{(𝑥-7{)}^3}\\ & \Rightarrow 2{𝑥}^2-3𝑥+5=𝐴(𝑥-7{)}^2+𝐵(𝑥-7)+𝐶\end{array}$

On putting $𝑥=7$ both sides, we get

$\begin{array}{cc}& (2\times 49)-(3\times 7)+5=0+0+𝐶\\ & \Rightarrow 𝐶=82\\ & \therefore 2{𝑥}^2-3𝑥+5=𝐴(𝑥-7{)}^2+𝐵(𝑥-7)+82\\ & \Rightarrow 2{𝑥}^2-3𝑥+5\\ & =𝐴{𝑥}^2-(14𝐴-𝐵)𝑥+7(7𝐴-𝐵)+82\end{array}$

On comparing both sides, we get

$\begin{array}{cc}& 𝐴=2,14𝐴-𝐵=3\\ \Rightarrow & 28-𝐵=3\\ \Rightarrow & 𝐵=25\\ \therefore & 2𝐴-3𝐵+𝐶=4-75+82=11\end{array}$

Given,

$\begin{array}{cc}& \frac{3{𝑥}^2+𝑎𝑥+3}{(2𝑥+3)({𝑥}^2+2)}=\frac{3}{2𝑥+3}+\frac{𝐵𝑥+𝐶}{{𝑥}^2+2}\\ & \Rightarrow 3{𝑥}^2+𝑎𝑥+3\\ & =3({𝑥}^2+2)+(𝐵𝑥+𝐶)(2𝑥+3)\\ & \Rightarrow 3{𝑥}^2+𝑎𝑥+3\\ & =3{𝑥}^2+6+2𝐵{𝑥}^2+3𝐵𝑥+2𝐶𝑥+3𝐶\\ & \Rightarrow 3{𝑥}^2+𝑎𝑥+3\\ & =(3+2𝐵){𝑥}^2+(3𝐵+2𝐶)𝑥+3𝐶+6\end{array}$

On comparing both sides, we get

$\begin{array}{cccc}3+2𝐵=3,& 3𝐶+6=3,& 3𝐵+2𝐶& =𝑎\\ \Rightarrow 𝐵=0& 𝐶=-1& 0-2& =𝑎\\ & & 𝑎& =-2\end{array}$

$\therefore 𝑎(𝐵+𝐶)=(-2)(0-1)=2$